# What is the pH of the solution which results from mixing 20.0mL of 0.50M HF(aq) and 50.0mL of 0.20M NaOH(aq) at 25 centigrades? (Ka of HF = 7.2 x 10^-4)

Apr 16, 2018

See below:

#### Explanation:

Let's start by finding the number of moles of $N a O H$ put into the solution, using the concentration formula:

$c = \frac{n}{v}$

$c$=conc in $m o l {\mathrm{dm}}^{-} 3$
$n$=number of moles
$v$=volume in litres ( ${\mathrm{dm}}^{3}$)

$50.0 m l = 0.05 {\mathrm{dm}}^{3} = v$

$0.2 \times 0.05 = n$

$n = 0.01 m o l$

And to find the number of moles of $H F$:

$c = \frac{n}{v}$

$0.5 = \frac{n}{0.02}$

$n = 0.1$

$N a O H \left(a q\right) + H F \left(a q\right) \to N a F \left(a q\right) + {H}_{2} O \left(l\right)$

We form 0.1 mol of $N a F$ in the resulting 70ml solution after the reaction has gone to completion.

Now, $N a F$ will be dissociated in the solution, and the fluoride ion, ${F}^{-}$ will act as a weak base in the solution(We will come back to this).

So now is a good time to set up an ICE table to find the amount of $O {H}^{-}$ ions it forms, but we first need to know the concentration of the $N a F$, as concentrations are used in the ICE table.

$c = \frac{n}{v}$

$c = \left(\frac{0.1}{0.07}\right)$

$c \approx 0.143 m o l {\mathrm{dm}}^{-} 3$ of $N a F$ ( =[F^(-)])

The reaction of the flouride ion and the consequent concentration changes are:

"color(white)(mmmmm)F^(-)(aq) + H_2O(l) -> HF(aq) + OH^(-)(aq)

$\text{Initial:} \textcolor{w h i t e}{m m} 0.143 \textcolor{w h i t e}{m m m} - \textcolor{w h i t e}{m m m m} 0 \textcolor{w h i t e}{m m m m l l} 0$

$\text{Change:} \textcolor{w h i t e}{i i m} - x \textcolor{w h i t e}{m m m} - \textcolor{w h i t e}{m m m} + x \textcolor{w h i t e}{m m l l} + x$

$\text{Eq:} \textcolor{w h i t e}{m m m} 0.143 - x \textcolor{w h i t e}{m i i} - \textcolor{w h i t e}{m m m m} x \textcolor{w h i t e}{m m m m l l} x$

The ${K}_{b}$ expression for the fluoride ion would be:

${K}_{b} = \frac{\left[O {H}^{-}\right] \times \left[H F\right]}{\left[{F}^{-}\right]}$

But how do we know the ${K}_{b}$ for the fluoride ion, which we touched upon earlier?

Well, as we are given that the reaction occurs at 25 degrees Celcius the following property applies:

$\left({K}_{b}\right) \times \left({K}_{a}\right) = 1.0 \times {10}^{-} 14$

For an acid/base pair- and we happen to have the pair of $H F$ and ${F}^{-}$!

Hence:

K_b=(1.0 times 10^-14)/(7.2 times 10^(-4)

${K}_{b} \left({F}^{-}\right) \approx 1.39 \times {10}^{- 11}$

So now we can create a ${K}_{b}$ expression and solve for $x$ to find the concentration of $O {H}^{-}$, and thus find the $p O H$ and then consequently the $p H$.

$1.39 \times {10}^{- 11} = \frac{{x}^{2}}{0.143 - x}$

${K}_{b}$ is small, so the small x approximation gives:

$1.39 \times {10}^{- 11} = \frac{{x}^{2}}{0.143}$

$x = \left[O {H}^{-}\right] = \sqrt{{K}_{b} \cdot 0.143}$

$= 1.4095 \times {10}^{- 6} \approx 1.41 \times {10}^{- 6}$

Now:

$p O H = - \log \left[O {H}^{-}\right]$
$p O H = - \log \left[1.41 \times {10}^{- 6}\right]$
$p O H \approx 5.85$

And as we are at 25 degrees this property applies:

$p H + p O H = 14$

Hence,

$p H = 14 - 5.85$

$\textcolor{b l u e}{p H = 8.15}$